Multidimensional array indexing question

조회 수: 1 (최근 30일)
Jason Nicholson
Jason Nicholson 2016년 8월 8일
편집: Jason Nicholson 2016년 8월 9일
I have a matrix x that is of size [61 2 45].
linearIndex = find(x(:,1,:) < x(:,2,:));
xAverage = (x(:,1,:) + x(:,2,:))/2;
Now I want to assign the average to anywhere x(:,1,:) < x(:,2,:). I come up with the following but it seems a bit verbose and un-elegant. Thoughts on how to do this better?
[subScriptIndex1, subScriptIndex2, subScriptIndex3] = ind2sub(size(linearIndex), linearIndex);
x(subScriptIndex1, 1, subScriptIndex3) = xAverage(subScriptIndex1, 1, subScriptIndex3);
x(subScriptIndex1, 2, subScriptIndex3) = xAverage(subScriptIndex1, 1, subScriptIndex3);
  댓글 수: 1
Jason Nicholson
Jason Nicholson 2016년 8월 8일
The new code looks like this:
linearIndex = linearIndex(:,[1,1],:);
x(linearIndex) = xAverage(linearIndex);
This is cleaner.

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채택된 답변

Stephen23
Stephen23 2016년 8월 8일
편집: Stephen23 2016년 8월 8일
Your understanding is correct: if a logical index is shorter than the array it is being used on, then the index is not expanded in any way. The solution is to make the index the exact size that you require:
x = reshape((1:18)',[3 2 3])
xx = x;
idx = x(:,1,:) < x(:,2,:);
idx = idx(:,[1,1],:) % or repmat
x(idx) = nan
[xx(:) x(:)]
  댓글 수: 3
Stephen23
Stephen23 2016년 8월 8일
Note that this behavior is closely related:
>> X = [1,2,3,4];
>> X([false,true]) % shorter than X
ans =
2
>> X([false,true,false(1,200)]) % longer than X, but only false..
ans =
2
Jason Nicholson
Jason Nicholson 2016년 8월 9일
Interesting.

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추가 답변 (1개)

Fangjun Jiang
Fangjun Jiang 2016년 8월 8일
x=rand(6,2,4);
MeanX=mean(x,2);
idx=x(:,1,:) < x(:,2,:);
x(idx)=MeanX(idx);
  댓글 수: 3
Fangjun Jiang
Fangjun Jiang 2016년 8월 8일
How about this? I think it works but what if the second dimension is larger than 2? There must be a a better way.
x=rand(6,2,4);
MeanX=mean(x,2);
MeanX(:,2,:)=MeanX(:,1,:);
idx=x(:,1,:) < x(:,2,:);
idx(:,2,:)=idx(:,1,:);
x(idx)=MeanX(idx);
Jason Nicholson
Jason Nicholson 2016년 8월 9일
편집: Jason Nicholson 2016년 8월 9일
Your second suggestion does work.
When the dimension is greater than 2, I think repmat may be the solution.

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